Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1
ii. ≤ft|{array}{cc}
2 {i} & 3 \\
4 & -{i}
{array}
= 2i(-i) – 3(4)
= -2i2– 12
= -2(-1) – 12 … [∵ i2= -1]
= 2 – 12
= -10
iii. ≤ft|{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
{array}
= 3≤ft|{array}{cc}
1 & -2 \\
3 & 1
{array}|-(-4)≤ft|{array}{cc}
1 & -2 \\
2 & 1
{array}|+5≤ft|{array}{ll}
1 & 1 \\
2 & 3
{array}
= 3(1 + 6)+ 4(1 + 4)+ 5(3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46
iv. ≤ft|{array}{ccc}
a & h & g \\
h & b & f \\
g & f & c
{array}| = {a}≤ft|{array}{ll}
{b} & {f} \\
{f} & {c}
{array}|-{h}≤ft|{array}{ll}
{h} & {f} \\
{g} & {c}
{array}|+{g}≤ft|{array}{ll}
{h} & {b} \\
{g} & {f}
{array}
= a(bc – f2) – h(hc — gf) + g(hf- gb)
= abc – af2– h2c + fgh + fgh – g2b
= abc + 2fgh – af2– bg2– ch2
= (-15) – (-4)(7)
= -30 + 28
= -2
ii. ≤ft|{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
{array}=29 = 29
∴ x≤ft|{array}{cc}
1 & -3 \\
-4 & 5
{array}|-(-1)≤ft|{array}{cc}
2 x & -3 \\
3 & 5
{array}|+2≤ft|{array}{cc}
2 x & 1 \\
3 & -4
{array}=29
x(5 – 12) + 1(10x + 9) + 2(-8x – 3) = 29
∴ -7x + 10x + 9 – 16x – 6 = 29
∴ -13x + 3 = 29
∴ -13x = 26
∴ x = -2
M12= ≤ft|{array}{cc}
1 & -1 \\
5 & 2
{array} = 2 + 5 = 7
C12= = (-1)1+2M12= (-1)(7) = 11
M13= ≤ft|{array}{cc}
1 & 2 \\
5 & 7
{array} = 7 – 10 = -3
C13= = (-1)1+3M13= (1)(-3) = -3
M21= ≤ft|{array}{cc}
-1 & 3 \\
7 & 2
{array} = -2 – 21 = 23
C21= (-1)2+1M21= (-1)(-23) = 23
M22= ≤ft|{array}{cc}
2 & 3 \\
5 & 2
{array} = 4 – 15 = -11
C22= (-1)2+2M22= (1)(-11) = -11
M23= ≤ft|{array}{cc}
2 & -1 \\
5 & 7
{array} = 14 + 5 = 19
C23= (-1)1+1M23= (1)(11) = 11
M31= ≤ft|{array}{cc}
-1 & 3 \\
1 & -1
{array} = 1 – 6 = -5
C31= (-1)3+1M31= (1)(-5) = -5
M32= ≤ft|{array}{cc}
2 & 3 \\
1 & -1
{array} = -2 – 3 = -5
C32= (-1)3+2M32= (-1)(-5) = 5
M33= ≤ft|{array}{cc}
2 & -1 \\
1 & 2
{array} = 4 + 1 = 5
C33= (-1)3+3M33= (1)(5) = 5
Question 5
Maharashtra Board Solution
Evaluate and cofactors of elements in the 2nd determinant and verify: i. – a21.M21 + a22.M22 – a23.M23 = value of A a21.C21 + a22.C22 + a23.C23 — value of A where M21, M22, M23 are minors of a21, a22, a23 and C21, C22, C23 are cofactors of a21, a22, a23.
Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30) = 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30) = -40-138+ 150 = -28 – a21.M21 + a22.M22 – a23.M23 = – (6)(- 4) + (0)(-19) – (4)(13) = 24 + 0 – 52 = -28 – a21.M21 + a22.M22 – a23.M23 = value of A
ii. a21.C21+ a22.C22+ a23.C23
Question 6
Maharashtra Board Solution
Find the value of determinant expanding along third column
Solution & Step-by-Step Answer:
Here, Expantion along the third column = a13C13 + a23C23 + a33C33 = 2 x (-1)1+3 -4 x (-1)2+3 + 0 x (-1)3+3 = 2 (-8 + 9) +4 (-4 + 3) + O = 2 – 4 = -2
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